Finding higher order Derivatives

  • Thread starter tmlrlz
  • Start date
  • #1
29
0

Homework Statement


Let p be an arbitrary polynomial
p(x) = anxn + an-1xn-1 + ... + a1x + a0, an cannot equal 0.
a) Find (dn/dxn)[p(x)]
b)What is (dk/dxk)[p(x)] for k>n


Homework Equations





The Attempt at a Solution


Im actually not really sure what to do for this question.

For the first part i tried to take the first three derivatives and got

(d/dx)[p(x)] = a1
(d2/dx2)[p(x)] = 2a2x + a1
(d3/dx3)[p(x)] = 3a3x2 + 2a2x + a1

(dn/dxn)[p(x)] = nanxn-1 + (n-1)a(n-1)xn-2 + ... + a1

Thats my guess, im not sure of how to do it.
As for part b:
I dont even know how to approach this question, my only assumption is that its the same answer again with all the n's replaced with k's. I am certain this is wrong so please help me do this question both a and b. Thank you.
 

Answers and Replies

  • #2
I like Serena
Homework Helper
6,577
176
Welcome to PF, tmlrlz! :smile:

What you have calculated is d/dx in each case.
dn/dxn means that you take the derivative n times.
 
  • #3
29
0
Welcome to PF, tmlrlz! :smile:

What you have calculated is d/dx in each case.
dn/dxn means that you take the derivative n times.
I thought i did that for part a, did i do it wrong? And how would you go about doing part b, what is it asking and how is it different from part a? Thank you.
 
  • #4
I like Serena
Homework Helper
6,577
176
I thought i did that for part a, did i do it wrong? And how would you go about doing part b, what is it asking and how is it different from part a? Thank you.
Let me give you an example:

You wrote:
(d2/dx2)[p(x)] = 2a2x + a1

But what you actually did was:
(d/dx)[p(x)] = 2a2x + a1

It should be:
(d2/dx2)[p(x)] = 2a2

Can you do this for say n=3?



If you understand what is intended for part a, I think you'll see what is intended for part b.
 
  • #5
118
0
The difference is that the n in the first part is the degree of the polynomial. The k in the second part is just any integer greater than the degree of the polynomial.
 
  • #6
I like Serena
Homework Helper
6,577
176
The difference is that the n in the first part is the degree of the polynomial. The k in the second part is just any integer greater than the degree of the polynomial.
Yes, this is true.
So I guess you answered the last part of your own question.
 
  • #7
29
0
Let me give you an example:

You wrote:
(d2/dx2)[p(x)] = 2a2x + a1

But what you actually did was:
(d/dx)[p(x)] = 2a2x + a1

It should be:
(d2/dx2)[p(x)] = 2a2

Can you do this for say n=3?


If you understand what is intended for part a, I think you'll see what is intended for part b.

oh i can't believe that i forgot to do the rest of the derivative!
so for n = 3
(d3/dx3)[p(x)] = 6a3

And then you can find part a
(dn/dxn)[p(x)] = n! * (an)

However i am still confused as to what part b is requiring?
 
  • #8
I like Serena
Homework Helper
6,577
176
oh i can't believe that i forgot to do the rest of the derivative!
so for n = 3
(d3/dx3)[p(x)] = 6a3

And then you can find part a
(dn/dxn)[p(x)] = n! * (an)

However i am still confused as to what part b is requiring?
Right! :smile:

Okay, let's pick a k>n.
I'll pick k=n+1.

What is (dn+1/dxn+1)[p(x)]?
 
  • #9
29
0
Right! :smile:

Okay, let's pick a k>n.
I'll pick k=n+1.

What is (dn+1/dxn+1)[p(x)]?
So then that means that when n = 1 we are taking the derivative of d2/dx2 right?

I took the first 3 derivatives again
when n =1
d2/dx2[p(x)] = d/dx (a1) = 0
when n =2
d3/dx3[p(x)] = 0
when n =3
d4/dx4[p(x)] = 0

so that means that
(dk/dxk)[p(x)] = 0

right? i hope!
 
Last edited by a moderator:
  • #10
I like Serena
Homework Helper
6,577
176
So then that means that when n = 1 we are taking the derivative of d2/dx2 right?

I took the first 3 derivatives again
when n =1
d2/dx2[p(x)] = d/dx (a1) = 0
when n =2
d3/dx3[p(x)] = 0
when n =3
d4/dx4[p(x)] = 0

so that means that
(dk/dxk)[p(x)] = 0

right? i hope!
Yep! :wink:

What do you think will happen if we pick k=n+2?
Or k=n+3?
Or any k>n?
 
  • #11
29
0
Yep! :wink:

What do you think will happen if we pick k=n+2?
Or k=n+3?
Or any k>n?
Then you will keep taking the derivative of zero which is zero over and over again till the end of the derivative instead of the previous case where you only got zero once, instead it will occur multiple times for that specific derivative
 
  • #12
I like Serena
Homework Helper
6,577
176
So? Any other questions?
 

Related Threads on Finding higher order Derivatives

Replies
9
Views
10K
  • Last Post
Replies
8
Views
7K
Replies
2
Views
915
  • Last Post
Replies
6
Views
2K
Replies
2
Views
3K
Replies
4
Views
1K
Replies
3
Views
879
Replies
1
Views
2K
Replies
5
Views
10K
Replies
10
Views
1K
Top